Test: Waves (CBSE Level With Solutions)


30 Questions MCQ Test Physics Class 11 | Test: Waves (CBSE Level With Solutions)


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This mock test of Test: Waves (CBSE Level With Solutions) for Class 11 helps you for every Class 11 entrance exam. This contains 30 Multiple Choice Questions for Class 11 Test: Waves (CBSE Level With Solutions) (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Waves (CBSE Level With Solutions) quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test: Waves (CBSE Level With Solutions) exercise for a better result in the exam. You can find other Test: Waves (CBSE Level With Solutions) extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

In wave propagation

Solution:

Because in wave propagation particles only moves in perodic motion, particles get collide with their neighbouring particles then transfer energy to them and comes back to their normal stage. Now next particle which gain energy get into excited state and starts moving periodically and get collide to next adjacent particle and so on. Thus in wave propagation particles only transfer their kinetic energy and momentum. Hence particles does not move therefore there is no flow of matter but there is movement of disturbance. 

QUESTION: 2

The speed of propagation of a sinusoidal wave is given by V=νλ where

Solution:

For a sinusoidal wave, V = v λ.
V = speed,
v = frequency,
λ = wavelength,
frequency (v) = reciprocal of the time period i.e. v =1/T

QUESTION: 3

Velocity of sound in air is 300 m/s. Then the distance between two successive nodes of a stationary wave of frequency 1000 Hz is.

Solution:

Velocity of sound in air= 300m/s =300×100=30000 cm/s
And frequency = 1000 hz
So, wavelength = Velocity/frequency
= 30000/1000= 30
Distance = wavelength/2
=30/2 = 15

QUESTION: 4

Two waves of wavelength 1m & 1.01 m produce 10 beats in 3 sec. The velocity of sound in a gas is about

Solution:

As we know, wavelength = speed/frequency
Wavelength 1 = 1m
Wavelength 2 = 1.01m
=> 1 = v/f1
f1 =  v
And 1.01 = v/f2
f2 = v/1.01
Now beat is 19 beats/3 secs
f1 - f2= 10/3
Solving equations we get
v = 336.6m/s
Hence B is the correct answer.

QUESTION: 5

In open end organ pipe

Solution:

In open end organ pipe both odd and even harmonics are produced. This is the definition of open end organ pipe.

QUESTION: 6

When we make a mobile telephone call to a friend

Solution:

Explanation:Because mobile communication is a space communication and in space communication basically electromagnetic waves are used (as carrier waves as in case of radio communication) because of the modulation ( frequency, amplitude) operations which can be performed on EM waves. Thus when our friend receives the call, he also receives EM waves which is the carrier of our audio signals.

QUESTION: 7

When two or more waves traverse the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is known as

Solution:
QUESTION: 8

There are three sources of sound of equal intensities with frequencies 400, 401 and 402 Hz. The number of beats per seconds is

Solution:

Resultant displacement of the wave by these three wave is
y=asin2π400t+asin2π401t+asin2π402t
y=a(1+2cos2πt)sin2π401t
So the resultant magnitude a(1+2cos2πt) has a maximum when,
cos2πt=1
or, t=0,1,2...
The time interval between two successive maximum is 1 sec.
So beat frequency is 1sec.

QUESTION: 9

Longitudinal waves cannot be propagated through

Solution:

Because longitudinal waves are the mechanical waves that need a medium to propagate such as air, gas, solid etc. but these are not available in vacuum, so this wave can't propagate in vacuum.

QUESTION: 10

A man is watching two trains one leaving and other approaching with equal velocities of 4 m/s. If they sound their whistles each of natural frequency 240 Hz, the number of beats heard per sec by the man will be (velocity of sound in air 320 m/s)

Solution:

QUESTION: 11

Electromagnetic waves are different from sound waves in that

Solution:

Explanation:Electromagnetic waves are transverse waves, they move perpendicular to the direction of propagation of wave ( the direction in which energy is transferred) and EM waves( Electromagnetic waves) can travel in vacuum, thus doesn't require any medium also.

QUESTION: 12

Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φφ, the result is a single wave with the same frequency ω:If φ = 0 or an integral multiple of 2 π

Solution:

Let the waves be
y1 = Asin(wt)
y2 = Asin(wt + φ)
If φ = 0 or 2nπ then the waves are exactly in phase and the interference is constructive.
Hence A is the correct answer.

QUESTION: 13

The velocity of sound in gas in which two waves of wavelength 1.0 m and 0.01 m produces 4 beats/sec. is

Solution:

Beats = |f1 – f2| where f1 and f2 are the frequencies
Now, f1 = v/λ1 and f2 = v/λ2
So, 4 = |v (1/1- 1/0.01)|
v = 0.0404 m/s

QUESTION: 14

Matter waves are useful in

Solution:

Explanation:Matter waves are also termed as De Broglie waves because they were initially introduced by him. All matters behave like a wave (during motion) and in there propagation there is action of forces due to collision between the particles, some time torque also exists, hence all the matters waves were consider to be a helpful part for understanding quantum.

QUESTION: 15

A fork of unknown frequency when sounded with another fork of frequency 256 Hz produces 4 beats/sec. The first fork is loaded with wax. It again produces 4 beats/sec. When sounded together with the fork of 256 Hz frequency, then the frequency of first tuning fork is

Solution:

After waxing the frequency of the tuning fork decreases and so the initial frequency must be higher so as to decrease it below 256 Hz in order to give 4 beats/sec.

QUESTION: 16

Speed of sound in air is 350 m/s. An engine blows a whistle of frequency of 1200 Hz, it is approaching the observer with velocity 50 m/s. The apparent frequency as heard by the observer is

Solution:
QUESTION: 17

The quantity similar to extension or compression of the spring in sound wave propagation (air) is

Solution:

Explanation:As in air wave propagates in the form of compression (increase in density of air) and rarefaction (decrease in density of air).

QUESTION: 18

When sound travels from air to water the quantity that remains unchanged is

Solution:

Because, frequency of a wave depend on source
V1/λ1 =V2/λ2 as frequency is constant
V1 - velocity of sound wave in air
V2 - velocity of sound wave in water

QUESTION: 19

56 tuning forks are arranged in a series that each fork gives 4 beats/sec, with the previous one. The frequency of last fork is 3 times that of first. Then frequency of first fork is

Solution:

Turning fork = 56turning
Beats number=4beats/s
A frequency of the last fork
Number of beats in each second= difference between frequency=4Hz
The frequency of the tuning fork of 56Hz beats3×n
(When n be the frequency of the first beat of the tuning fork)
Now, There are 55 difference between a set of 56 tunning fork.
The difference between two frequencies4Hz
Now, the difference between 56Hz4 and 1st tunning frequencies=3n−n
= 2n
As per equation,
3n−n=2n
2n=55×4
n = 55×4/2
= 110Hz

QUESTION: 20

Two persons cannot hear each other on the surface of moon because the moon has

Solution:
QUESTION: 21

Travelling or progressive wave

Solution:

Explanation:As progressive wave means a wave propagating in some onward direction in the medium

QUESTION: 22

A motor cyclist moving at 30 km/hr blows a whistle of 476 Hz towards a cliff. If velocity of sound is 1220 km/hr, the apparent frequency of the echo heard by him is

Solution:
QUESTION: 23

An echo repeats two syllables. If the velocity of sound is 330 m/s, then the distance of reflecting surface is

Solution:

Let us say that we speak syllables at a rate of 2 to 9 per second. So let us say that a syllable takes a minimum of 0.1 sec for a fast speaker. Let us say that a sound pulse (syllable) is emitted starting at t = 0.

The effect of a syllable lasts on the ear for 0.1 sec. So if any echo reaches the year before t = 0.2 sec., then it is mixed with the direct sound present in the ear and so echo is not properly heard.

In this problem, two syllables are repeated in the echo. That is it took about 2 * 0.2 sec ie., 0.4 seconds for the sound to travel to the reflecting surface and come back to the ear.

The distance of the reflecting surface from the person

= 330 m/s * 0.4 sec / 2

= 66 meters.

QUESTION: 24

The waves on the surface of water are of two kinds:.

Solution:

Explanation:

As capillary & gravity waves are elastic waves or mechanical waves which require medium for their propagation.

Hence they are using the elastic behaviour of water.

Hence

The waves on the surface of water are of two kinds:

capillary waves and gravity waves

QUESTION: 25

A certain broadcasting station broadcasts at frequency of 12 Mega Hz. The wavelength of waves emitted is

Solution:

c = λν
Therefore, λ = 25 m

QUESTION: 26

A tuning fork produces 4 beats/sec. with 50 cm and 40 cm of a stretched wire, of a sonometer. The frequency of fork is

Solution:
QUESTION: 27

When sound travels from air to water the quantity that remains unchanged is

Solution:
QUESTION: 28

Which one is not produced by sound waves in air?

Solution:

Polarisation is a phenomenon of converting unpolarized light to polarized. It is not done by sound waves.

QUESTION: 29

in the same medium transverse and longitudinal waves

Solution:

Explanation:

As speed of transverse & longitudinal waves depend on different modulus of elasticity (Young's modulus, Bulk modulus, Modulus of Rigidity) of the medium.

So in the same medium transverse and longitudinal waves

travel with different speeds

QUESTION: 30

If a star emitting orange light moves away from the earth, its color will

Solution:

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum (towards the orange/red/infrared/microwave/radio end of the spectrum).